Review of Acid and Base Definition

• Arrhenius Definition

( ) OHOH + H l2

-

(aq)(aq) ←

→+

• ACIDS – donates H+

• HNO3, H3PO4, H2SO4, HCl, HI, HBr,

CH3COOH, organic-COOH, H2SO3

• BASES – donates OH-

• NaOH, KOH, LiOH, CsOH,

Mg(OH)2,Sr(OH)2,Ba(OH)2 , Ca(OH)2

Acids and Bases II:

• Bronsted-Lowry Definition

++⇔+ 4(aq)

-

(aq)3(g)(l)2 NHOH NH OH

• ACIDS – donates H+

• (proton donor)

• HNO3, H3PO4, H2SO4, HCl, HI, HBr,

CH3COOH, organic-COOH, H2SO3

• BASES – accepts H+

• (proton acceptor)

• NH3, organic-NH2, NaOH, KOH, LiOH,

CsOH, Mg(OH)2,Sr(OH)2,Ba(OH)2 , Ca(OH)2

• Bronsted-Lowry Definition: Correct assignment of acid-base

conjugate pairs is based on properly drawn Lewis structures

• Amphoterism:

– Species that can behave as an acid or base are

called amphoteric.

– Amphiprotic: special term for an amphoteric

species which involves proton transfer reactions

to show behavior as either an acid or base

OH 2 ) Zn(NO HNO 2 Zn(OH) 22332 +→+

-2

4

-

2 Zn(OH)OH 2 Zn(OH) →+

HPO42- + H2O ���� H2PO4

- + OH-

HPO42- + H2O ���� PO4

3- + H3O+

• Lewis Definition

• ACIDS – electron-pair acceptor

• H+ (∴ all molecules with H+)

• Electron deficient molecules (below-octet

atoms eg. Boron cmpds)

• BASES – electron-pair donor

• OH- (∴ all molecules with OH-)

• Molecules with lone e- pairs

acid base

OH NH OH NH -

423 ← +→+ +

acid base

BF Na BF NaF -

43 +←

→+ +

Acids and Bases: Chem 16 � 17

• The Autoionization of water

← +→+ + -

(aq)(aq)3(l)2(l)2 OHOHOH OH

Equilibrium-constant expression:2

2

-

3c

]O[H

]][OHOH[K

+

=

But concentration of water is constant

(and large) at 25oC, therefore: ]][OHOH[K K -

3wc

+=⇒

Experimental concentration H+ is

determined to be 1.00x10-7 at 25oC,

therefore: C25at 101.00x K

)10x 00.1)(10(1.00x K

o14-

w

-7-7

w

=

=

pH = -log [H3O+] or simply –log[H+]

pOH = -log[OH-]

Kw = 1.00 x 10-14 = [H+][OH-] at 25oC

∴ pOH + pH = 14.00

Proof:

-log Kw = -log [1.00 x 10-14] = -log ([H+][OH-])

-log Kw = 14.00 = -log[H+] + -log[OH]-

14.00 = pH + pOH

• Calculate the [H3O+], pH, [OH-], and pOH for a 0.020 M HNO3

solution.

– Is HNO3 a weak or strong acid?

– What is the [H3O+] ?

[ ]( )

70.1pH

100.2-logpH

100.2OH

0.020 0.020 020.0

NOOHOHHNO

2

2

3

-

33

100%

23

=

×=

×=

+ →+

−

−+

+≈

M

M

MMM

• What is the pH of water at its normal boiling point? Is it acidic or basic?– Given:

∆Hfo

H2O(l) = -285.83 kJ/mol

∆Hfo

OH-(aq) = -230.0 kJ/mol

� NOTE: When concentration of H+ coming from an acid source

(e.g. HA) reaches 1.00 x 10-5 and below, the [H+] of H2O should

be added, where [H+] = 1.00 x 10-7 (only used at 25oC)

� Example: What is the pH of a solution prepared by diluting

1.0 mL of 0.1 M HCl with 1000 liters of water?

� The weaker the acid or base, the stronger the conjugate partner.

� The stronger the acid or base, the weaker the conjugate partner.

2(g)(l)2(aq)32(aq)3(aq)3(aq)32

(aq)3(aq)(aq)(aq)3

COOHCOHOO2NaCHOOHCH2CONa

COOHCH NaCl HCl OONaCH

+⇒+→+

+→+

STRONGER

ACID and BASE

WEAKER

ACID and BASE

Relative Strengths of Acids and Bases

Conjugate Acid-Base Pairs

Ionization Constants for Monoprotic

Weak Acids and Bases

• Consider an aqueous solution of acetic acid,

CH3COOH. What is the equilibrium constant

expression?

CH3COOH(aq) + H2O(l) ⇄ CH3COO-(aq) + H3O+

(aq)

]OCOOH][H[CH

]COO][CHOH[K

23

-

33c

+

=

But [H2O] = 55.6 M, very high and almost constant, therefore

COOH][CH

]COO][CHOH[KK

3

-

33

ac

+

=≈

• We can simply write the equation for

dissociation as

CH3COOH(aq) ⇄ CH3COO-(aq) + H+

(aq)

COOH][CH

]COO][CHH[K

3

-

3a

+

=

COOH][CH

]COO][CHOH[KK

3

-

33

ac

+

=≈

Ka is the acid-dissociation constant

• For weak bases,

NH3(aq) + H2O(l) ⇄ NH4+

(aq) + OH-(aq)

O]][H[NH

]][NHOH[K

23

4c

+−

=][NH

]][NHOH[K

3

4b

+−

=

Kb is the base-dissociation constant.

Does the base really dissociate, like acids?

[acid]

]base e][conjugatH[Ka

+

=

[base]

]acid e][conjugatOH[Kb

−

=

HA ⇄ H⇄ H⇄ H⇄ H++++ + A+ A+ A+ A----

B- + H2O ⇄ OH⇄ OH⇄ OH⇄ OH---- + BH+ BH+ BH+ BH

� The ionization constant values for several acids are given below.

◦ Which acid is the strongest?

Acid Formula Ka value

Acetic CH3COOH 1.8 x 10-5

Nitrous HNO2 4.5 x 10-4

Hydrofluoric HF 7.2 x 10-4

Hypochlorous HClO 3.5 x 10-8

Hydrocyanic HCN 4.0 x 10-10

Ionization Constants for Monoprotic Weak Acids

and Bases

� The order of decreasing acid strength for these weak acids is:

HCN>HClO>COOHCH>HNO>HF 32

� Calculate the concentrations of the various species in 0.15 M acetic acid, CH3COOH, solution.

� ALWAYS write down the ionization reaction and the ionization constant expression.

[ ][ ][ ]

5

3

-

33a

-

3323

108.1COOHCH

COOCHOHK

COOCHOH OHCOOHCH

−+

+

×==

← +→+

xMxM-x)M.(

xMxMxM

M

++

++

← +→+ +

150 ] [ mEquilibriu

- Change

0.15 ] [ Initial

COOCH OH OHCOOHCH -

3323

• Short-cut: Use the simplfying assumption-

Since x << 0.15, assume that 0.15 – x ≈ 0.15≈ 0.15≈ 0.15≈ 0.15

If ��

�������

1.0 � 10��, the simplifying assumption is valid

Percent Ionization of Weak

Acids/Bases

� Calculate the percent ionization of 0.15 M acetic acid. The percent ionization of acetic acid is

[ ] [ ][ ]

%1.1%10015.0

106.1ionization %

%100COOHCH

Hor COOCH= ionization %

3

original3

-

3

=××

=

×

−

+

M

M

equilequil

Percent Ionization of Weak

Acids/Bases• Calculate the percent ionization of 0.15 M

hydrocyanic acid, HCN. Ka = 4.0 x 10-10

– Compare the %ionization of HCN and HOAc.

� Note that the [H+] (or %ionization) in 0.15 M acetic acid is 215 times greater than for 0.15 M HCN.

Solution Ka [H+] pH % ionization

0.15 M HOAc 1.8 x 10-5 1.6 x 10-3 2.80 1.1

0.15 M HCN 4.0 x 10-10 7.7 x 10-6 5.11 0.0051

Solvolysis: Reaction of Acid/Base with solvent

• Solvolysis - reaction of a substance with the solvent in which it is dissolved.– Hydrolysis refers to the reaction of a substance with water

or its ions.

� Consider the acid HA

HA + H2O ⇄ A- + H3O+ Ka

Reverse form:

A- + H3O+ ⇄ HA + H2O Ka’ = 1/Ka

� Consider the conjugate base, A-

A- + H2O ⇄ HA + OH- Kb

Reverse form:

HA + OH- ⇄ A- + H2O Kb‘ = 1/Kb

Solvolysis: Reaction of Acid/Base with solvent

• How is Ka related to Kb?

HA + H2O ⇄ A- + H3O+ Ka

A- + H2O ⇄ HA + OH- Kb

H2O + H2O ⇄ H3O+ + OH- Kw

baw K x KK =a

wb

K

KK =

b

wa

K

KK =

� The order of decreasing acid strength for the weak acids is:

HCN>HClO>COOHCH>HNO>HF 32

� The order of increasing base strength of the anions

(conjugate bases) of the same acids is:

---

3

-

2

- CN<ClO<COOCH<NO<F

1.8 x 10-54.5 x 10-47.2 x 10-4 3.5 x 10-8 4.0 x 10-10

a

wb

K

KK =

5.6 x 10-101.4 x 10-112.2 x 10-11 2.9 x 10-7 2.5 x 10-5

The stronger the acid/base, the weaker is its conjugate

• In a 0.12 M solution of a weak monoprotic acid, HY, the acid is 5.0% ionized. Calculate the dissociation constant for the weak acid.

� The pH of a 0.10 M solution of a weak monoprotic

acid, HA, is found to be 2.97. What is the value of

its dissociation constant?

Strengths of Acids and Bases• Strengths of BINARY Acids - acid strength increases with

decreasing H-X bond strength.

– VIIA hydrohalic acids

Bond strength has this periodic trend

HF >> HCl > HBr > HI

Acid strength has the reverse trend.

HF << HCl < HBr < HI– VIA hydrides.

Bond strength has this trend.

H2O >> H2S > H2Se > H2Te

The acid strength is the reverse trend.

H2O << H2S < H2Se < H2Te

Down a group: ����size, ����energy to break H- bond

(����electronegativity), ����acidity

Strengths of Acids and Bases

Arrange in order of increasing acidity:

NH3, OH2, HF

���� NH3 < OH2 < HF

(Electronegativity trend: NH3 < OH2 < HF)

• Across a period: ����electronegativity, ����acidity

Strengths of Acids and Bases

• TERNARY ACIDS - hydroxides of nonmetals

that produce H3O+ in water.

– Consist of H, O, and a nonmetal.

HClO4 H3PO4

Strengths of Acids and Bases

• Acidity of ternary acids with same central

element increase with increasing oxidation state

of central element, and increasing O atoms

HClO < HClO2 < HClO3 < HClO4

weakest strongest

Cl oxidation states

+1 +3 +5 +7

Strengths of Acids and Bases

• ternary acids of the same group and same number of O atoms increase in acidity with increase electronegativity of central atom

H2SeO4 < H2SO4

HBrO4 < HClO4

HBrO3 < HClO3

However for phosphorus

ternary acids:

H3PO2 > H3PO3 > H3PO4 –

relative position of H is

important (based on

structures)

Base Strength of Amines

� The electronic properties of

the substituents (alkyl groups

enhance the basicity, aryl

groups diminish it).

� Steric hindrance offered by

the groups on nitrogen.

Polyprotic Acids/Bases• Many weak acids contain two or more acidic hydrogens.

– Examples include H3PO4 and H3AsO4.

• The calculation of equilibria for polyprotic acids is done in a stepwise fashion.

– There is a dissociation constant for each step

• Consider arsenic acid, H3AsO4, which has three ionization constants.

1 Ka1 = 2.5 x 10-4

2 Ka2 = 5.6 x 10-8

3 Ka3 = 3.0 x 10-13

• Calculate the concentration of all species in 0.100 M

arsenic acid, H3AsO4, solution.

You may apply the simplifying assumption in each step (1 ICE table/

dissociation)

11-

a2(aq)3

-2

3(aq)(l)2

-

3(aq)

-7

a1(aq)3

-

3(aq)(l)23(aq)2

10 x 4.7 K OH COOH HCO

10 x 4.4K OH + HCOOHCOH

=+↔+

=↔++

+

8-

b2(aq)3(aq)2(l)2

-

3(aq)

-4

b1(aq)

-

3(aq)(l)2

-2

3(aq)

10 x 2.3 K HO COHOH HCO

10 x 1.2K HO + HCOOHCO

=+↔+

=↔+−

−

b2a2

w

b1a1

KK

K

KK

K HO POHOH POH

K HO POHOH HPO

K HO + HPOOHPO

b3(aq)4(aq)3(l)2

-

4(aq)2

b2(aq)

-

4(aq)2(l)2

-2

4(aq)

b1(aq)

-2

4(aq)(l)2

-3

4(aq)

−

−

−

+→+

+→+

→+

b3a3

b2wa2

b1a1

KK

KKK

KK

13-

a3(aq)3

-3

4(aq)(l)2

-2

4(aq)

8-

a2(aq)3

-2

4(aq)(l)2

-

4(aq)2

-3

a1(aq)3

-

4(aq)2(l)24(aq)3

10 x 3.60 K OH POOH HPO

10 x 6.20 K OH HPOOH POH

10 x 50.7K OH + POHOHPOH

=+→+

=+→+

=→+

+

+

+

• The behaviour of an amphiprotic species (acting as base or acid) depends on its dissociation constants

11-

a2(aq)3

-2

3(aq)(l)2

-

3(aq)

-8

b2(aq)3(aq)2(l)2

-

3(aq)

10 x 4.80 K OH COOH HCO

10 x 2.38 K OH COHOH HCO

=+→+

=+→++

−

12-

b3(aq)4(aq)3(l)2

-

4(aq)2

-8

a2(aq)3

-2

4(aq)(l)2

-

4(aq)2

10 x 1.33 K OH POHOH POH

10 x 6.20 K OH HPOOH POH

=+→+

=+→+−

+

7-

b2(aq)

-

4(aq)2(l)2

-2

4(aq)

-13

a3(aq)3

-3

4(aq)(l)2

-2

4(aq)

10 x 1.61 K OH POHOH HPO

10 x 3.6 K OH POOH HPO

=+→+

=+→+−

+

• What is the pH of the resulting solution obtained by dissolving 1.52 g of NaH2PO4•2H2O in 50.00 mL water? If the salt added was Na2HPO4•2H2O instead, will the solution be basic or acidic?

• You ‘accidentally’ spilled muratic acid (2.0 M HCl) on the rubber flip-flops of your roommate. To neutralize the acid, you looked for a base in the dorm stock room, and you found two salts –sodium bicarbonate and sodium phosphate. Which of the two salts will you use to quickly and effectively neutralize the acid?

Inorganic Lewis Acids – Hydrolysis of

Metal Ions• Because metal ions are positively charged, they

attract the electrons of oxygen atoms in water.

– Depending on the strength of electron interacting with the cation, the water molecule can turn into hydroxide anion and release H+

– The acid strength of these ion-complexes acting as Lewis acids depend on size and charge of cation center

3-

a(aq)

2

(aq)52

3

)aq(62 10 x 2.0 K H (OH)O)Fe(H )OFe(H =+←

→ +++

Na(NO3) Ca(NO3)2 Zn(NO3)2 Al(NO3)3

7.0 6.9 5.5 3.5

Salts of acids and bases� Aqueous solutions of salts of strong acids and strong bases

are neutralExamples: NaCl (from HCl and NaOH)

K2SO4 (from KOH and H2SO4)

� Aqueous solutions of salts of strong bases and weak acidsare basic

Examples: NaCN (from NaOH and HCN)

K2C2O4 (from KOH and H2C2O4)

� Aqueous solutions of salts of weak bases and strong acidsare acidic

Examples: NH4Cl (from NH3 and HCl)

(CH3)3NHBr ((CH3)N and HBr)

How about – KHC2O4? NaHSO4? LiHSO3?

Salts of acids and bases

• Aqueous solutions of salts of weak bases and weak acids can be neutral, basic or acidic.

The values of Ka and Kb determine the pH.

NH4CH3COO?

Compare Ka of NH4+ vs Kb of OAc-

MgNH4PO4?

Compare Ka of NH4+ and Mg2+ vs Kb of OAc-

NH4(HCO3)?

Compare Ka of NH4+ vs Kb/Ka of amphiprotic HCO3

-

Common Ion Effect

and

Buffers/Buffer Capacity

Common Ion Effect –

A special name for a Le Chatelier-based shift

• Consider a solution of 0.05 M acetic acid, CH3COOH (50.00

mL)

5-

a(aq)3(aq)3)l(2(aq)3 10 x 1.8 K OH COOCH OH COOHCH =+←

→+ +−

OH COOCH OH COOHCH (aq)3(aq)3)l(2(aq)3

+− +←+

OH COOCH OH COOHCH (aq)3(aq)3)l(2(aq)3

+− +←+

• Describe the direction of equilibrium shift

After adding 10.00 mL of 0.5 M HCl

After adding 10.00 mL of 0.5 M NaCH3COO

In COMMON-ION effect, the direction of shift of equilibrium is always towards the

side that diminishes the added common/similar ion

Common Ion Effect –

A special name for a Le Chatelier-based shift

• Consider a solution of 0.05 M acetic acid,

CH3COOH (50.00 mL)

• Describe the pH of the final mixture

– After adding 10.00 mL of 0.5 M HCl

– After adding 10.00 mL of 0.5 M NaCH3COO

DECREASE pH, more ACIDIC

INCREASE pH, less ACIDIC

Buffers

• Solutions that contain BOTH acid component and

its conjugate base

– Conjugate base is present in the initial concentration

of components

– Examples :

• Acetic acid added with sodium acetate

• Ammonium chloride added with aqueous ammonia solution

• Solutions that resist drastic pH changes

Henderson-Hasselbalch Equation• Simplified equation for pH calculation involving

buffers

Henderson-Hasselbalch Equation• What is pH of a buffer that is 0.12 M in lactic acid, HC3H5O5,

and 0.10 M in sodium lactate?

Ka = 1.40 x 10-4

The most important aspect of buffer solutions is that they resist drastic

changes of pH upon adding strong acids or bases!

OH COOOHC OH COOHOHC (aq)3(aq)344)l(2(aq)344

+− +←

→+

0.10 M0.12 M

+ x + x - x

I

C

E 0.12 - x 0.10 + x x

(0.12)

(0.10) pK pH a +=

x) (0.12

x) (x)(0.10 K a −

+=

Henderson-Hasselbalch Equation

HA + H2O ⇄ A⇄ A⇄ A⇄ A---- + H+ H+ H+ H3333OOOO++++

B + H2O ⇄ BH⇄ BH⇄ BH⇄ BH++++ + HO+ HO+ HO+ HO----

� HH equation only used when salt is present – that is, present separately, not the

[salt] from ICE calculation

� The salt-component must be added separately, or generated by neutralizing the

main component

� In calculations involving buffers, ICE table must be in terms of MOLES especially if

volumes are not same. However, equating with Ka must be in MOLARITY.

� HH equation is allowed only when “simplifying assumptions” are valid

Preparation of Buffers• Buffers can be prepared in three ways

– Adding a solid salt component to a liquid• Ex. NaCH3COO solid added to a solution of acetic acid (acetic-

acetate buffer)

– Neutralizing a liquid component with a strong opposite component • Ex. Aqueous ammonia added with liquid HCl (ammonia-

ammonium buffer)

Aqueous phosphoric acid added with solid NaOH (phosphate buffer)

– Mixing two solid salts in the same volume of water• Ex. Solid NaH2PO4•H2O and solid Na2HPO4•7H2O dissolved in

water (phosphate buffer)

Preparation of Buffers

• How many grams of NH4Cl must be added to 2.0 L

of 0.10 M NH3 to form a buffer of pH 9.00?

Kb NH3 = 1.8 x 10-5

� What volume of 0.5 M NaOH must be added to 50 mL

of 0.1 M benzoic acid (C6H5COOH) to make 100.0 mL of

0.05 M benzoate buffer that is pH 4.5?

Ka benzoic acid = 6.3 x 10-5

Preparation of Buffers• Prepare a 100.0 mL 0.1 M phosphate buffer, pH 8.00.

– Given:

H3PO4

pKa1 = 2.12

pKa2 = 7.21

pKa3 = 12.38

Molarity H3PO4 (liquid) = 14.85 M

Formula weight

NaH2PO4•H2O = 137.99 g/mole

Na2HPO4•7H2O = 268.07 g/mole

7.21 pK OH HPOOH POH a2(aq)3

-2

4(aq)(l)2

-

4(aq)2 =+→+ +

)POH (moles

)HPO (moleslog pK pH

-

42

-2

4a2 +=

componentsbuffer moles totalHPO moles POH moles -2

4

-

42 =+

M) (0.1)

L 1mL 1000(

mL) (100.0 HPO moles POH moles -2

4

-

42 ×=+

Common Ion Effect – Buffers

• Non-buffer Case: Consider the a solution of 50.00 mL of 0.05 M acetic acid, CH3COOH. – Determine the pH of the solution.

– Calculate the pH of the final mixture after adding 10.00 mL of 0.05 M HCl.

– What is the ∆pH?

OH COOCH OH COOHCH (aq)3(aq)3)l(2(aq)3

+− +←

→+

� Because HCl is a strong acid, it directly contributes to the initial concentration of H3O

+ in the equilibrium calculation� Set-up ICE table in MOLE basis, then convert to molarity when calculating/equating to Ka.

(10.00 mL) •(0.05 M)

(50.00 mL) •(0.05 M)

+ x + x + x + x + x + x + x + x ---- x x x x

IIII

CCCC

• Buffer CASE - Consider a solution of 50.00 mL 0.05 M acetic acid

(CH3COOH) and 0.05 M sodium acetate (NaCH3COO).

– Determine the pH of the solution.

– Calculate the pH of the final mixture after adding 10.00 mL of 0.05 M HCl.

– What is the ∆pH?

Step 1: STOICHIOMETRIC calculation

Which has the lowest change in pH (∆pH)?

� Try calculating the pH after adding 10.00 mL of 0.05 M NaOH (instead HCl) for the 2 cases.

Step 2: EQUILIBRIUM calculation, HH

mL) 60.00 volume(total

M) mL)(0.05 (60.00

mL) 60.00 volumetotal(

M) mL)(0.05 00.40(

log pKpH a

=

=

+=

Buffer Capacity

�Amount of acid or base (usually in mL) needed to

change the pH of a buffer solution by 1 degree.

� ∆B – number of moles of acid/base added per 1 L of buffer solution

� ∆pH – change in pH associated with the addition of acid/base

� ↑∆B, ↓ ∆pH, ↑β,

Compare the two buffers:

- 100 mL of 1.0 M NaCH3COO and 1.0 M CH3COOH

- 100 mL of 0.1 M NaCH3COO and 0.1 M CH3COOH

-Which has the highest buffer capacity relative to 1.0 M NaOH?

Buffer capacity is highest when pH = pKa

Acid-Base Neutralizations:

Indicators, Titrations

and

pH curves

Acid-Base Indicators

• The point at which chemically equivalent amounts of acid and

base have reacted is called the equivalence point.

• The point at which a chemical indicator changes color is called

the end point.

2color 1color

OH In OH HIn (aq)3(aq))l(2(aq)

+−→← ++

bcolor acolor

OH HIn OH In (aq)(aq))l(2(aq)

−−→← ++

Acidic indicator

Basic indicator

Acid-Base Indicators

• The equilibrium constant expression for an indicator

would be expressed as:

[HIn]

]][InO[H K

-

3In

+

= ][In

[HIn]K ]O[H

-In3 =+

pH range when indicator

changes its color depends

largely on Ka of the indicator

Acid-Base Indicators

Color change ranges of some acid-base indicators

Indicator

Color in

acidic

range pH range

Color in

basic range

Methyl violet Yellow 0 - 2 Purple

Methyl orange Pink 3.1 – 4.4 Yellow

Litmus Red 4.7 – 8.2 Blue

Phenolphthalein Colorless 8.3 – 10.0 Red

Titration Curves

• Strong Acid titrated with Strong Base

Given: 25.00 mL of 0.5 M HClO4, calculate the pH of the resulting

solution after adding the following volumes of 0.5 M NaOH:

Volume of

0.5 M NaOH

(mL)

Mmoles

NaOH

Mmoles

HClO4

remaining

Total

Volume

(mL)

[H+]final pH

0.00 0.0 12.5 25.0 0.5000 0.301

5.00 2.5 10.0 30.0 0.3333 0.477

10.00 5.0 7.5 35.0 0.2143 0.669

15.00 7.5 5.0 40.0 0.1250 0.903

20.00 10.0 2.5 45.0 0.0555 1.256

25.00 12.5 0 50.0 1x10-7 7.000

30.00 15.0 0 55.0 2.2x10-13 12.657

Methyl orange

Phenolphthalein

Litmus

Methyl violet

• Strong Acid/ Strong Base Titration Curve

0

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pH

Volume Titrant (0.5 M NaOH)

SA-SB curve

Titration Curves• Weak Acid titrated with Strong Base

Given: 25.00 mL of 0.5 M CH3COOH, calculate the pH of the resulting

solution after adding the following volumes of 0.5 M NaOH:

Volume of

0.5 M

NaOH (mL)

Mmoles

NaOH

Mmoles

CH3COOH

remaining

Mmoles

CH3 COO-

produced

Total

Volume

(mL)

pH

0.00 0.0 12.5 0 25.0 0.301

5.00 2.5 10.0 2.5 30.0 4.143

10.00 5.0 7.5 5.0 35.0 4.569

15.00 7.5 5.0 7.5 40.0 4.921

20.00 10.0 2.5 10.0 45.0 5.348

25.00 12.5 0 12.5 50.0 9.071

30.00 15.0 0 12.5 55.0 12.658

Methyl orange

Phenolphthalein

Litmus

Methyl violet0

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0 5 10 15 20 25 30 35 40

pH

Titrant volume

WA-SB curve

WA vs SB Titration Curve Regions

0

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pH

Titrant volume

WA-SB curve

Initial Region: Use ICE

and Ka

pH = ½ (pKa + pCHA)

Buffer Region: HH Equation

pH = pKa + log (moles A- /moles HA)

At half equivalence point:

pH = pKa

Equivalence Region: Use ICE

and Kb

pOH =

½ (pKb + p[moles HA/total

volume])

Excess base region:

pOH = -log (moles excess base/ total volume)

Buffer Region: HH Equation

pH = pKa + log [moles titrant/(moles analyte –

moles titrant )]

Methyl orange

Phenolphthalein

Litmus

Methyl violet

0

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acetic curve

lactic curve

HClO curve

Titration

Curve of

Different

Acids vs

Strong Base

• A 0.1044-g sample of an unknown monoprotic acid required

22.10 mL of 0.0500 M NaOH to reach the endpoint. (a) What

is the molecular weight of the acid? (b) As the acid is titrated,

the pH of the solution after the addition of 11.05 mL of the

base is 4.89. What is the Ka of the acid?

� A biochemist needs 750 mL of an acetic acid-sodium acetate

buffer with pH 4.50. Solid sodium acetate, NaC2H3O2, and

glacial acetic acid, HC2H3O2, are available. Glacial acetic acid is

99% pure by mass and has a density of 1.05 g/mL. If the

buffer is to be 0.20 M in HC2H3O2, how many grams of the salt

and how many milliliters of glacial acetic acid must be used?

� What is the pH of a solution made by mixing 0.30 mole

NaOH, 0.25 mole Na2HPO4 , and 0.20 mole H3PO4 with

water and diluting with 1.00 L?

H3PO4

pKa1 = 2.12

pKa2 = 7.21

pKa3 = 12.38